Recent content by mathgal

So I have: F'(x)= 1 if 0<x<1 0 if 1<x<2 -2x+4 if x>2 Like that?

So for my formula for F', is there something specific I have to write for x=1 and x=2? How would I do that since my intervals are 0<= x <= 1 and 1<x<2 and x>=2 where there's ones and twos overlapping?

But doesn't the Fundamental Theorem of Calculus tell us that since our F is continuous everywhere, that it should be differentiable there? How do I show that x=1,2 are not differentiable? I know if the limit from the left and right are equal, it is continuous at that point.

I think I actually understood that better than I thought and was just getting confused by my own writing. Thank you so much. I know you've already helped a lot, but do you have any ideas about how I would then find a formula for F'(x) wherever F is differentiable? My test is tomorrow, that is...

F(x)= x if 0 <= x <= 1. Then, for 1< x <2, we're taking the integral of 0, so I should just get a constant? I just don't know how I am supposed to write that part of the third part whre x>=2

okay well thanks, anyways

Oh and in Post #11, I meant to say: F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1

Oh and in Post #11, I meant to say: F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 1

But we're evaluating it from (0,1)... So once we integrate and get x|, then we have 1-0=1. This isn't correct?

Actually shouldn't F(x)= 1, 0<=x<=1, F(x)=1, 1<x<2, and F(x)=-x^2+4x-3, x>=2

so should F(x)=x also if 1<x<2? F(x)= int (1,2)= int(0,1) 1 dx + int (1,2) 0 dx = x + 0 = 0 is that correct?

Also, is there a way for me a find a formula for F'? How do I first find where F is differentiable?

Well it should actually be F(x)=-x^2+4x-4 for x>= 2 right?

Is this correct? I'm worried the last one should be more complicated? F(x) = x, 0 <= x <= 1 = 0, 1 < x < 2 = -x^2+4x, x >= 2.

Should I be using limits? I actually did graph it. I know what the F(x) should be, I'm just not sure how to rigorously show it using a theorem, formula, etc